Đáp án:

a. +∞

Giải thích các bước giải:

 Bài 1:

\(\begin{array}{*{20}{l}}
{a.\mathop {\lim }\limits_{x \to {\rm{ \;}} - \infty } {x^2}\left( {1 - \frac{{12}}{x} + \frac{{12}}{{{x^2}}}} \right) = {\rm{ \;}} + \infty }\\
{Do:\mathop {\lim }\limits_{x \to {\rm{ \;}} - \infty } {x^2} = {\rm{ \;}} + \infty }\\
{\mathop {\lim }\limits_{x \to {\rm{ \;}} - \infty } \left( {1 - \frac{{12}}{x} + \frac{{12}}{{{x^2}}}} \right) = 1}\\
{b.\mathop {\lim }\limits_{x \to {\rm{ \;}} - 1} \frac{{12{x^2} + 15x + 3}}{{x + 1}}}\\
{ \to \mathop {\lim }\limits_{x \to {\rm{ \;}} - 1} \frac{{3.\left( {4x + 1} \right)\left( {x + 1} \right)}}{{x + 1}}}\\
{ = \mathop {\lim }\limits_{x \to {\rm{ \;}} - 1} 3.\left( {4x + 1} \right) = 3\left[ {4.\left( { - 1} \right) + 1} \right] =  - 9}\\
{c.\mathop {\lim }\limits_{x \to {\rm{ \;}} - \infty } \frac{{12{x^2} + 15x + 3}}{{{x^2} - 1}}}\\
{ = \mathop {\lim }\limits_{x \to {\rm{ \;}} - \infty } \frac{{12 + \frac{{15}}{x} + \frac{3}{{{x^2}}}}}{{1 - \frac{1}{{{x^2}}}}} = \frac{{12}}{1} = 12}
\end{array}\)

Bài 2:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = \mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} + 13x + 12}}{{x + 1}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{x\left( {x + 1} \right) + 12\left( {x + 1} \right)}}{{x + 1}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 1} \right)\left( {x + 12} \right)}}{{x + 1}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \left( {x + 12} \right)\\
 = 12 - 1 = 11\\
f\left( { - 1} \right) = 12 - 1 = 11\\
 \to f\left( { - 1} \right) = \mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = 11
\end{array}\)

⇒ Hàm số liên tục tại x=-1